Is there a way to compute this antiderivative by u-substitution?

Question

I started by trying to find an expression for $\int \ln(ax^n + b) dx$ where $a,n,b$ are integers $>0$. So naturally I used integration by parts and arrived at $$x\ln(ax^n + b) - \int \dfrac{ax^{n}}{ax^n + b}dx$$ Is there a way to compute $\int \frac{ax^{n}}{ax^n + b}dx$ by u-substitution or any other method?

Answer 1

Hint:

In some certain cases depending on the powers, you may use Chebyshev theorem on the integration of binomial differentials for such that integral.

Answer 2

Solution in which I suppose $n > 0$

There is a way, but you shall know special functions, like the HyperGeometric Special Function.

Let's collect $ax^n$:

$$\int\frac{ax^n}{ax^n\left(1 + \left(\frac{b}{a}x^{-n}\right)\right)}\ \text{d}x$$

With the help of Geometric Series we find

$$\int\sum_{k = 0}^{+\infty} \left(-\frac{b}{a}x^{-n}\right)^k\ \text{d}x = \sum_{k = 0}^{+\infty}\left(-\frac{b}{a}\right)^k\int x^{-kn}\ \text{d}x$$

Integration is trivial, the resulting series is

$$\sum_{k = 0}^{+\infty}\left(-\frac{b}{a}\right)^k \frac{x^{-kn + 1}}{-kn+1}$$

This series does converge to the HyperGeometric Special Function:

$$\sum_{k = 0}^{+\infty}\left(-\frac{b}{a}\right)^k \frac{x^{-kn + 1}}{-kn+1} = x\cdot _2F_1\left[1, -\frac{1}{n}, 1 - \frac{1}{n}, -\frac{b}{a}x^{-n}\right]$$

More on HSF

https://en.wikipedia.org/wiki/Hypergeometric_function

Answer 3

Outline: With a change of variable we end up integrating $\ln(c(t^n+1))$ for some positive constant $c$, and then we only need to find $\int \ln(t^n+1)\,dt$. It is too early to integrate by parts. Instead, factor $t^n+1$ over the reals.

Depending on whether $n$ is odd or even, we get that $t^n+1$ is $t+1$ times a product of irreducible quadratics, or just a plain product of irreducible quadratics. These irreducible quadratics are all of the shape $t^2-(2\cos\theta)t +1$. The logarithm of this product is a sum of logarithms.

We look at the case $n$ even. The case $n$ odd is similar. So we are integrating terms of the shape $\ln(t^2-(2\cos\theta) t+1)$. The integration by parts procedure suggested in the OP works. It is helpful to first complete the square.